polynomial function (original) (raw)

Remark. The coefficients ai in a polynomial function need not be unique; e.g. if R={0, 1} is the ring (and field) of two elements, then the polynomials X and X2 both may be used for the same polynomial function. However, if we stipulate that R is an infinite integral domain, the coefficients are guaranteed to be unique.

The set of all polynomial functions of R, being a subset of the set RR of all functions from R to R, is here denoted by R/R.

Theorem.

If R is a commutative ring, then the set R/R of all polynomial functions of R, equipped with the operations

is a commutative ring.

Proof. It’s straightforward to show that the function set RR forms a commutative ring when equipped with the operations “+” and “⋅” defined as (1). We show now that R/R forms a subring of RR. Let f and g be any two polynomial functions given by

Then we can give f+g by

where k=max⁡{m,n} and ai=0 (resp. bi=0) for i>m (resp. i>n). This means that f+g∈R/R. Secondly, the equation

signifies that f⋅g∈R/R. Because also the function -f given by

and satisfying -f+f=0:x↦0 belongs to R/R, the subsetR/R is a subring of RR.